Unit 1 Topic 3: Exponential
Equations in One Variable
Algebra 1 Summit
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Printed: August 11, 2014
AUTHOR
Algebra 1 Summit
www.ck12.org
Chapter 1. Unit 1 Topic 3: Exponential Equations in One Variable
1
C HAPTER
Unit 1 Topic 3: Exponential
Equations in One Variable
Dear Parents/Guardians,
Unit 1 Topic 3 focuses on exponential equations in one variable. This topic includes:
• Using the Properties of Exponents to simplify exponential expressions
• Using the Exponential Property of Equality to solve exponential equations
Properties of Exponents:
Consider the following exponential expressions with the same base and what happens through the algebraic operations. Exponential expressions have several components, for example:
axb
a is the coefficient
x is the base
b is the exponent or power
Product Property (multiplication):
am × an = am+n
When the bases are the same then exponents can be added.
a)
32 × 33
The base is 3.
32+3
Keep the base of 3 and add the exponents.
3
5
This answer is in exponential form.
The answer can be taken one step further. The base is numerical so the term can be evaluated.
35 = 3 × 3 × 3 × 3 × 3
35 = 243
32 × 33 = 35 = 243
Note: Rules of exponents are shortcut methods for simplifying. When in doubt expand out the values as shown
below to simplify (x3 )(x2 )
(x • x • x)(x • x) = x5
expanding expressions works for all rules.
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b)
(x3 )(x6 )
The base is x.
x3+6
Keep the base of x and add the exponents.
x
9
The answer is in exponential form.
(x3 )(x6 ) = x9
c)
5x2 y3 · 3xy2
The bases are x and y.
15(x2 y3 )(xy2 )
Multiply the coefficients - 5 × 3 = 15. Keep the bases of x and y and add
the exponents of the same base. If a base does not have a written
exponent, it is understood as 1.
15x
2+1 3+2
y
3 5
15x y
The answer is in exponential form.
5x2 y3 · 3xy2 = 15x3 y5
e) Simplify the expression: x3 y2
this expression cannot be simplified further since the bases are not the same
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Chapter 1. Unit 1 Topic 3: Exponential Equations in One Variable
Quotient Property (division):
am
= am−n
an
The quotient property is an extension of the product property, but may end with the possibility of a negative in the
exponent. When dividing, if the bases are the same the exponents can be subtracted. If a negative exponent results
from subtraction, see the negative exponent rules below.
a)
27 ÷ 23
The base is 2.
2
7−3
Keep the base of 2 and subtract the exponents.
2
4
The answer is in exponential form.
The answer can be taken one step further. The base is numerical so the term can be evaluated.
24 = 2 × 2 × 2 × 2
24 = 16
27 ÷ 23 = 24 = 16
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b)
x8
x2
x8−2
x6
The base is x.
Keep the base of x and subtract the exponents.
The answer is in exponential form.
x8
= x6
x2
c)
y3
y−5
The base is y.
+
y3−−5
y
8
Keep the base of y and subtract the exponents.
The answer is in exponential form.
y3
= y8
y−5
d) Note: when simplifying all answers must be written with positive exponents, (see next section for explanation)
16x2 y5
4x5 y3
2 5
x y
4 5 3
x y
The bases are x and y.
Divide the coefficients - 16 ÷ 4 = 4. Keep the bases of x and y and
subtract the exponents of the same base.
4x
2−5 5−3
y
−3 2
4x y
All answers must be written with positive exponents. If the base is in the
numerator with a negative exponent, write it in the denominator with a
positive exponent.
4y2
x3
16x2 y5 4y2
= 3
4x5 y3
x
4
The answer is in exponential form.
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Chapter 1. Unit 1 Topic 3: Exponential Equations in One Variable
Negative exponent:
a−m =
1
am
A negative exponent means reciprocal. This is because of multiplying the base by "a" m times, you are dividing m
times.
a) x−3 =
1
x3
b) 4r−2 /t would simplify to 4/ r2 t
note: variable should be ordered alphabetically when simplifying
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Power to a Power:
(am )n = amn
a)
(23 )2
The base is 23 .
23×2
Keep the base of 23 and multiply the exponents.
26
The answer is in exponential form.
The answer can be taken one step further. The base is numerical so the term can be evaluated.
26 = 2 × 2 × 2 × 2 × 2 × 2
26 = 64
(23 )2 = 26 = 64
b) (a3 )4 would be simplified using the power to power rule (a 3x4 ) = a12
Justification: ((a · a · a)(a · a · a)(a · a · a)(a · a · a) = a12
The answer can be taken one step further. The base is numerical so the term can be evaluated.
c) (3y4 )2
Justification: 32 · (y4 )2
the two distributes to both parts of the inside of the parenthesis:
Answer: 9y8
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Chapter 1. Unit 1 Topic 3: Exponential Equations in One Variable
Zero Property:
The zero property of exponents states that anything raised to the zero property becomes 1.
a) 30 = 1
b) c0 = 1
Two videos by James Sousa:
First on Properties of Exponents: http://www.youtube.com/watch?v=0GAMbuPJGOY
Second on Negative Exponents: https://www.youtube.com/watch?v=R7Yp5TW1NTs
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More complex examples applying the properties of exponents:
Example A
Simplify the following expression. All exponents should be positive in the final answer.
(a−2 b3 )−3
ab2 c0
Solution:
(a−2 b3 )−3
a6 b−9
a5
=
=
ab2 c0
ab2 · 1 b11
Example B
Simplify the following expression until all exponents are positive.
(2x)5 ·
42
=
2−3
Solution:
note: the 42 was rewritten as
24 because they are equvalent
and then could be combined with the
other base 2’s.
(2x)5 ·
8
42
25 x5 24 29 x5
=
= −3 = 212 x5
2−3
2−3
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Chapter 1. Unit 1 Topic 3: Exponential Equations in One Variable
Example C
Simplify the following expression until all exponents are positive.
(a−3 b2 c4 )−1
(a2 b−4 c0 )3
Solution:
(a−3 b2 c4 )−1 a3 b−2 c−4
b10
=
=
(a2 b−4 c0 )3
a6 b−12
a3 c4
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Using the Exponential Property of Equality to solve exponential equations:
In this section, students will be able to construct arguments to justify their reasoning in problems involving exponential expressions and equations. In addition, students will be able to articulate the differences and similarities
in solving linear and exponential equations. In order to do this, they need to have a clear understanding of the
Exponential Property of Equality and be able to use it to find equivalent exponential expressions and solve equations
involving exponents.
The Exponential Property of Equality states that two exponential expressions with the same base will have equal
exponents.
If bx = by , then x = y .
Example A
If 2x = 25 , then x = 5.
Example B
Solve for x.
5x+1 = 53x
Apply the exponential
property of equality, when the bases
are the same, the exponent
can be set equal.
x + 1 = 3x
Apply Subtraction Property of Equality. x + 1 − x = 3x − x
1 = 2x
Apply Division Property of Equality.
1 ÷ 2 = 2x ÷ 2
1
Solution.
2 =x
Students will learn to apply this property to solve more complex exponential equations by finding common bases,
then setting the exponents equal and solving.
Video about solving equations using the exponential property of equality:
https://www.youtube.com/watch?v=aPyE9SKtczs
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Chapter 1. Unit 1 Topic 3: Exponential Equations in One Variable
Example C
Solve for n.
Since 27 = 33 , then
Simplify.
3n = 272
3n = (33 )2
3n = 36
n=6
Example D
Solve for x.
Rewrite in a common base.
Simplify.
Set exponents equal.
Apply Subtraction Property of Equality.
Solution: −1 = x
22x−1 = 8x
22x−1 = (23 )x
2x−1
2
= 23x
2x − 1 = 3x
2x − 1 − 2x = 3x − 2x
Example E
Solve for m.
1
25
= 1252m
Rewrite in a common base.
= (53 )2m
−2
Simplify.
5 = 56m
Set exponents equal.
−2 = 6m
Apply Division Property of Equality. −2 ÷ 6 = 6m ÷ 6
Simplify.
− 13 = m
5−2
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Example F
Solve for x.
1 x
1003x+1 = ( 1000
)
2(3x+1)
10
= (10−3 )x
6x+2
10
= 10−3x
6x + 2 = −3x
Rewrite in a common base.
Simplify.
Set exponents equal.
Apply Subtraction Property of Equality.
6x + 2 − 6x = −3x − 6x
2 = −9x
Apply Division Property of Equality.
2 ÷ −9 = −9x ÷ −9
− 92 = x
Example G
Janet and Isaac make cupcakes to sell at the school bake sale. They each make 1 cupcake to sample. The next day
Janet doubles her production. She continues to double it each day for the next few weeks. Six days after he make
the sample Isaac has gathered some friends to help. He makes eight times as many cupcakes than the day before
and continues in this pattern.
a) On what day will they have the same number of cupcakes?
b) How many cupcakes will they have on that day?
Solution:
a) Set up and solve an equation
2x = 8x−6
2x = 23(x−6)
x = 3x − 18
x=9
On day 9, Janet and Isaac will have the same number of cupcakes.
b) 29 = 512
On day 9, Janet and Isaac each have 512 cupcakes.
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